and henceIn [1]}, the choice \(\alpha _1=1/\xi \)  was made, which implies,
given the condition (REF ) found previously, that
\(\alpha _2=-1+1/\xi ^2\)  and hence the last term in (REF ) vanishes.
\(j(\varphi ) = {\square ^{-1}}\, j^\rho \circ \tilde{j}_\rho +\alpha _1\, {\square ^{-1}}\, j^\alpha (c)\circ \tilde{j}_\alpha ({{\tilde{c}}}) +[(1+\alpha _2)\, \xi ^2 -1]\, {\square ^{-2}}\, {\partial }j\circ {\partial }\tilde{j}\,.\) 
